[csu/coj 1078]多个序列的最长公共子序列

题意：给n个序列，同一个序列里面元素互不相同，求它们的最长公共子序列。

思路：任取一个序列，对于这个序列里面的两个数ai,aj(i<j)，如果对于其它每一个序列，都出现过ai,aj，且ai在aj之前出现，那么i到j连一条长度为1的有向边，完美转化为DAG最长路。需要注意：对于某个数，如果某个序列没出现那么这个点的答案应该为-INF，表示这个点表示的状态不合法。

代码：

  1 #pragma comment(linker, "/STACK:10240000,10240000")
  2
  3 #include <iostream>
  4 #include <cstdio>
  5 #include <algorithm>
  6 #include <cstdlib>
  7 #include <cstring>
  8 #include <map>
  9 #include <queue>
 10 #include <deque>
 11 #include <cmath>
 12 #include <vector>
 13 #include <ctime>
 14 #include <cctype>
 15 #include <stack>
 16 #include <set>
 17 #include <bitset>
 18 #include <functional>
 19 #include <numeric>
 20 #include <stdexcept>
 21 #include <utility>
 22
 23 using namespace std;
 24
 25 #define mem0(a) memset(a, 0, sizeof(a))
 26 #define mem_1(a) memset(a, -1, sizeof(a))
 27 #define lson l, m, rt << 1
 28 #define rson m + 1, r, rt << 1 | 1
 29 #define define_m int m = (l + r) >> 1
 30 #define rep_up0(a, b) for (int a = 0; a < (b); a++)
 31 #define rep_up1(a, b) for (int a = 1; a <= (b); a++)
 32 #define rep_down0(a, b) for (int a = b - 1; a >= 0; a--)
 33 #define rep_down1(a, b) for (int a = b; a > 0; a--)
 34 #define all(a) (a).begin(), (a).end()
 35 #define lowbit(x) ((x) & (-(x)))
 36 #define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {}
 37 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {}
 38 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {}
 39 #define pchr(a) putchar(a)
 40 #define pstr(a) printf("%s", a)
 41 #define sstr(a) scanf("%s", a)
 42 #define sint(a) scanf("%d", &a)
 43 #define sint2(a, b) scanf("%d%d", &a, &b)
 44 #define sint3(a, b, c) scanf("%d%d%d", &a, &b, &c)
 45 #define pint(a) printf("%d\n", a)
 46 #define test_print1(a) cout << "var1 = " << a << endl
 47 #define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl
 48 #define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = " << b << ", var3 = " << c << endl
 49 #define mp(a, b) make_pair(a, b)
 50 #define pb(a) push_back(a)
 51
 52 typedef long long LL;
 53 typedef pair<int, int> pii;
 54 typedef vector<int> vi;
 55
 56 const int dx[8] = {0, 0, -1, 1, 1, 1, -1, -1};
 57 const int dy[8] = {-1, 1, 0, 0, 1, -1, 1, -1 };
 58 const int maxn = 1e3 + 7;
 59 const int md = 1e9 + 7;
 60 const int inf = 1e9 + 7;
 61 const LL inf_L = 1e18 + 7;
 62 const double pi = acos(-1.0);
 63 const double eps = 1e-6;
 64
 65 template<class T>T gcd(T a, T b){return b==0?a:gcd(b,a%b);}
 66 template<class T>bool max_update(T &a,const T &b){if(b>a){a = b; return true;}return false;}
 67 template<class T>bool min_update(T &a,const T &b){if(b<a){a = b; return true;}return false;}
 68 template<class T>T condition(bool f, T a, T b){return f?a:b;}
 69 template<class T>void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];}
 70 int make_id(int x, int y, int n) { return x * n + y; }
 71
 72 struct Graph {
 73     vector<vector<int> > G;
 74     void clear() { G.clear(); }
 75     void resize(int n) { G.resize(n + 2); }
 76     void add(int u, int v) { G[u].push_back(v); }
 77     vector<int> & operator [] (int u) { return G[u]; }
 78 };
 79 Graph G;
 80
 81 int mp[12][12 * maxn];
 82 int dp[maxn], a[12][maxn], b[12 * maxn];
 83 bool vis[maxn];
 84
 85 int dfs(int pos) {
 86     if (vis[pos]) return dp[pos];
 87     vis[pos] = true;
 88     int sz = G[pos].size();
 89     rep_up0(i, sz) {
 90         max_update(dp[pos], dfs(G[pos][i]) + 1);
 91     }
 92     return dp[pos];
 93 }
 94
 95 int main() {
 96     //freopen("in.txt", "r", stdin);
 97     int n, m;
 98     while (cin >> n >> m) {
 99         rep_up0(i, n) {
100             rep_up0(j, m) {
101                 sint(a[i][j]);
102                 b[i * m + j] = a[i][j];
103             }
104         }
105         sort(b, b + n * m);
106         int sz = unique(b, b + n * m) - b;
107         mem0(mp);
108         rep_up0(i, n) {
109             rep_up0(j, m) {
110                 a[i][j] = lower_bound(b, b + sz, a[i][j]) - b;
111                 mp[i][a[i][j]] = j + 1;
112             }
113         }
114         G.clear();
115         G.resize(m);
116         rep_up0(i, m) dp[i] = 1;
117         rep_up0(i, m) {
118             rep_up1(j, n - 1) {
119                 if (!mp[j][a[0][i]]) {
120                     dp[i] = -inf;
121                     break;
122                 }
123             }
124         }
125         rep_up0(i, m) {
126             for (int j = i + 1; j < m; j ++) {
127                 bool ok = true;
128                 int x = a[0][i], y = a[0][j];
129                 rep_up1(k, n - 1) {
130                     if (!(mp[k][x] && mp[k][y] && mp[k][x] < mp[k][y])) {
131                         ok = false;
132                         break;
133                     }
134                 }
135                 if (ok) G[i].push_back(j);
136             }
137         }
138         int ans = 0;
139         mem0(vis);
140         rep_up0(i, m) max_update(ans, dfs(i));
141         cout << ans << endl;
142     }
143     return 0;
144 }